Question

If $\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }-\frac{\cos \theta }{\sqrt{1+{\cot }^2\theta }}-2\tan \theta \cot \theta =-1,\theta \in [0,2\pi ]$, then

• A. $\theta \in (0,\frac{\pi }{2})-\left\{\frac{\pi }{4}\right\}$
• B. $\theta \in (\frac{\pi }{2},\pi )-\left\{\frac{3\pi }{4}\right\}$
• C. $\theta \in (\pi ,\frac{3\pi }{2})-\left\{\frac{5\pi }{4}\right\}$
• D. $\theta \in (0,\pi )-\{\frac{\pi }{4},\frac{\pi }{2}\}$

Answer 1

Answer: (D)

$\theta \in (0,\pi )-\{\frac{\pi }{4},\frac{\pi }{2}\}$

We know that ${\sin }^3\theta -{\cos }^3\theta =(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )=(\sin \theta -\cos \theta )(1+\sin \theta \cos \theta )$

$\Rightarrow \frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }=1+\sin \theta \cos \theta$

But for it to exist $\sin \theta -\cos \theta \ne 0$ , which implies $\theta \ne \frac{\pi }{4}$

The value of $\frac{\cos \theta }{\sqrt{1+{\cot }^2\theta }}=|\sin \theta |\cos \theta$

So the given equation becomes $1+\sin \theta \cos \theta -|\sin \theta |\cos \theta -2\tan \theta \cot \theta =-1$

We know that $\tan \theta \cot \theta =1$, remember that for $\theta =\frac{\pi }{2}$, $\tan \theta$ does not exist.

So the given equation reduces to $\sin \theta \cos \theta =|\sin \theta |\cos \theta$

$\Rightarrow |\sin \theta |=\sin \theta$

Which implies $0<\theta <\pi$ and $\theta \ne \frac{\pi }{4},\frac{\pi }{2}$

So the correct option is $D$